If 0 < α , β < 2 π , then the number of ordered pairs ( α , β ) satisfying sin 2 | TestHub

Passage / Comprehension

$y = a x^{2} + b x + c = 0$ , for all $a, b, c \in R$ with $a \neq = 0$ , is a quadratic equation which has real roots if and only if $b^{2} - 4 a c \geq 0$ . If $F (x, y) = 0$ is a second-degree equation, then using the above fact, we can determine the range of $x$ and $y$ by treating it as a quadratic equation in $y$ or $x$ . Similarly, $a x^{2} + b x + c \geq 0$ for all $x \in R$ if $a > 0$ and $b^{2} - 4 a c \leq 0$ .

If $0 < α, β < 2 π$ , then the number of ordered pairs $(α, β)$ satisfying

$sin^{2} (α + β) - 2 sin α sin (α + β) + sin^{2} α + cos^{2} β = 0$

Options:

Answer:

Solution:

Solving it, we get $sin (α + β) = sin α \pm - cos^{2} β$

$\Rightarrow cos β = 0 \Rightarrow β = \frac{π}{2}, \frac{3 π}{2}$

(i) If $β = \frac{π}{2} \Rightarrow tan α = 1, α \in {\frac{π}{4}, \frac{5 π}{4}}$

(ii) If $β = \frac{3 π}{2} \Rightarrow tan α = - 1, α \in {\frac{3 π}{4}, \frac{7 π}{4}}$

Mathematics - Quadratic Equation Question with Solution | TestHub

Options:

Doubts & Discussion

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